Leetcode|递归分治法|面试题 08.06. 汉诺塔问题

1 递归分支法

用数学归纳法解决

• n = 1，盘子从A直接移动到C
• n = 2
• n = k，将A上n-1个盘子移动到B，然后将最底下的盘子移动到C，最后将B中的盘子移动到C

代码语言：javascript复制

class Solution {
public:
    void hanota(vector<int>& A, vector<int>& B, vector<int>& C) {
        int n = A.size();
        move(n, A, B, C);
    }

    void move(int n, vector<int>& A, vector<int>& B, vector<int>& C) {
        if (n == 1) {
            C.push_back(A.back());
            A.pop_back();
        } else {
            // 将柱子A上面n-1个盘子从A移动到B
            move(n - 1, A, C, B);
            // 将柱子A底下最大的盘子移动到柱子C
            C.push_back(A.back());
            A.pop_back();
            // 将柱子B上面n-1个盘子从B移动到C
            move(n - 1, B, A, C);
        }
    }
};

2 输出打印盘子移动过程（快手广告算法二面题）

当n = 3时

代码语言：javascript复制

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;

void move(int n, vector<int>& A, vector<int>& B, vector<int>& C, string s) {
    if (n == 1) {
        cout << "盘子" << A.back() << " from " << s[0] << " to " << s[2] << endl; 
        C.push_back(A.back());
        A.pop_back();
        return;
    }
    swap(s[1], s[2]);
    move(n - 1, A, C, B, s);
    cout << "盘子" << A.back() << " from " << s[0] << " to " << s[1] << endl; 
    C.push_back(A.back());
    A.pop_back();
    swap(s[1], s[2]);
    swap(s[0], s[1]);
    move(n - 1, B, A, C, s);
}

int main() {
    int n = 3;
    vector<int> A, B, C;
    for (int i = n; i > 0; i--)
        A.emplace_back(i);
    move(n, A, B, C, "ABC");
	return 0;
}

递归 算法

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